Power series with "shifted" central binomial coefficients During a tedious calculation, I arrived at a power series of the form which I want to compute:
$$ \sum\limits_{n=m}^\infty r^{2n} \binom{2n}{n-m}  $$
But from the search I so far did in other threads, I saw only ad-hoc methods whose conditions I don't think suit here. My current attempts have not been fruitful, but  I hope someone would perhaps nudge me in the right direction on how to deal with this.
 A: We find in H. Wilf's Generatingfunctionology formula (2.5.15) providing a generating function of the shifted  central binomial  coefficients in the form
\begin{align*}
\frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^k=\sum_{n}\binom{2n+k}{n}x^n\tag{1}
\end{align*}

We can adapt (1) easily and obtain
  \begin{align*}
\color{blue}{\sum\limits_{n=m}^\infty}&\color{blue}{ \binom{2n}{n-m}r^{2n}}\\
&=\sum_{n=0}^\infty\binom{2n+2m}{n}r^{2n+2m}\\
&=r^{2m}\sum_{n=0}^\infty\binom{2n+2m}{n}\left(r^2\right)^n\\
&\,\,\color{blue}{=\frac{1}{\sqrt{1-4r^2}}\left(\frac{1-\sqrt{1-4r^2}}{2r}\right)^{2m}}
\end{align*}

A: This answer is  based upon the Lagrange inversion theorem. We follow the paper Lagrange Inversion: when and how by R. Sprugnoli etal. It is convenient to apply formula G6 from the paper, which states that if there are functions $F(u)$ and $\phi(u)$, so that
\begin{align*}
C_n=[u^n]F(u)\phi(u)^n\tag{1}
\end{align*}
the following is valid
\begin{align*}
f(r)&=\sum_{n=0}^{\infty}C_nr^n\\
&=\sum_{n=0}^{\infty}[u^n]F(u)\phi(u)^nr^n\\
&=\left.\frac{F(u)}{1-r\phi^{\prime}(u)}\right|_{u=r\phi(u)}\tag{2}
\end{align*}
$[u^n]$ is the coefficient of operator  denoting the coefficient of $u^n$ in a series.

We obtain
  \begin{align*}
\color{blue}{\sum_{n=m}^\infty r^{2n}\binom{2n}{n-m}}&=\sum_{n=0}^\infty\binom{2n+2m}{n}r^{2n+2m}\tag{3}\\
&=r^{2m}\sum_{n=0}^\infty[u^n](1+u)^{2n+2m}r^{2n}\tag{4}\\
&=r^{2m}\left.\frac{(1+u)^{2m}}{1-r^2\cdot 2(1+u)}\right|_{u=r^2(1+u)^2}\tag{5}\\
&=r^{2m}\frac{(1+u)^{2m+1}}{1-u}\tag{6}\\
\end{align*}

Comment:


*

*In (3) we shift the index to start with $n=0$.

*In (4) we apply the coefficient of operator $[u^n](1+u)^{2n+2m}=\binom{2n+2m}{n}$.

*In (5) we apply (2) with $F(u)=(1+u)^{2m}, \phi(u)=(1+u)^2$ evaluated at $r^2$.

*In (6) we substitute $r^2=\frac{u}{(1+u)^2}$ and do some simplifications.
The next step is to solve the quadratic equation $u=r^2(1+u)^2$ (see (5)) and take the solution $u=u(r)$ which can be expanded as generating function around $0$. We obtain
\begin{align*}
u_{0,1}(r)=\frac{1}{2r^2}\left(1\pm\sqrt{1-4r^2}\right)-1
\end{align*}
We take the solution $u_0=\frac{1}{2r^2}\left(1\color{blue}{-}\sqrt{1-4r^2}\right)-1$ and obtain
\begin{align*}
1+u&=\frac{1}{2r^2}\left(1-\sqrt{1-4r^2}\right)\\
1-u&=2-\frac{1}{2r^2}\left(1-\sqrt{1-4r^2}\right)\\
&=\frac{\sqrt{1-4r^2}}{2r^2}\left(1-\sqrt{1-4r^2}\right)
\end{align*}

Substituting the results for $1+u$ and $1-u$ in (6) we obtain
  \begin{align*}
\color{blue}{\sum_{n=m}r^{2n}\binom{2n}{n-m}}&=r^{2m}\frac{(1+u)^{2m+1}}{1-u}\\
&=\frac{r^{2m}}{\sqrt{1-4r^2}}\left(\frac{1-\sqrt{1-4r^2}}{2r^2}\right)^{2m}\\
&\,\,\color{blue}{=\frac{1}{\sqrt{1-4r^2}}\left(\frac{1-\sqrt{1-4r^2}}{2r}\right)^{2m}}
\end{align*}

